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Continuity of a function

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    f:R2 -> R

    f(x,y) = [tex]e^{-x^{2}/y^{2}}[/tex] if y is not 0, and 0 if y is 0

    a) At (1,1), is f continuous?
    b) At (1,0), is f continuous?

    2. Relevant equations

    The function f is continuous at the point c if for every sequence (xn) in X with limit lim xn = c, we have lim f(xn) = f(c).

    3. The attempt at a solution

    a) I think it is continuous at (1,1). My reasoning is that [tex]e^{-x^{2}/y^{2}}[/tex] is continuous on [tex]R^{2}-\left\{(x,0)\right\}[/tex], and [tex](1,1) \in R^{2}-\left\{(x,0)\right\}[/tex].

    b) I don't think this is true... but not sure how to show it.

    I am trying to work with the definition: "The function f is continuous at the point c if for every sequence (xn) in X with limit lim xn = c, we have lim f(xn) = f(c)."

    So if f is continuous at (1,0), that means for any lim (xn,yn) = (1,0), we have lim f(xn,yn) = f(1,0). But f(1,0) is 0, so any lim f(xn,yn) = 0.

    If I choose (xn,yn) = (1, 1/n), the limit is (1,0). But lim f(xn,yn) = lim [tex]e^{-n^{2}}[/tex], which does not equal 0. Does this show that f is not continuous at point (1,0)?
     
    Last edited: Nov 17, 2009
  2. jcsd
  3. Nov 16, 2009 #2

    Mark44

    Staff: Mentor

    For b, if your function is continuous at (1, 0), then lim f(x, y) = f(1, 0) = 0, as (x, y) approaches (1, 0), independent of the path taken to get to (1, 0). If you can find one path into (1, 0) that produces a function value other than 0, that would suffice to show that the limit doesn't exist.
     
  4. Nov 17, 2009 #3
    Is my example of such a path (that I stated above) not valid?

    If (xn, yn) = (1, 1/n), it goes to (1,0), but the limit of the function value, lim f(xn, yn) = lim [tex]e^{-1/\frac{1}{n^{2}}}[/tex] = lim [tex]e^{-n^{2}}[/tex], which does not equal 0... it goes to infinity as n goes to infinity.
     
    Last edited: Nov 17, 2009
  5. Nov 18, 2009 #4
    okay, I guess it wasn't. f IS continuous at (1, 0), and the function value of any path into (1, 0) is always 0.

    For f to be continuous at (1,0), this condition must hold: [tex]lim_{(x, y)\rightarrow(1, 0)} f(x,y) = f(1,0)=0[/tex]

    First, if [tex]y=0[/tex], then f(x, y)=f(1, 0)=0. Second, if [tex]y\neq0[/tex], since [tex](x,y) \rightarrow (1,0)[/tex], we have [tex]-x^{2}/y^{2} \rightarrow -\infty[/tex] and thus [tex]f(x,y)=e^{-\infty} \rightarrow 0[/tex]. So f must be continuous at (1, 0)
     
  6. Nov 18, 2009 #5

    Mark44

    Staff: Mentor

    Yes, I agree. In your previous post, you said that lim f(xn, yn) = lim e-n2, which does not equal zero (limit as n --> infinity for yn = 1/n). As n gets large w/o bound, though, e-n2 approaches 0.
     
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